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Spiel & Spaß
  
Thema eröffnet von: mire2
MP-Stilblüten etc. sammeln  
Beitrag No.1392 im Thread
Zum letzten BeitragZum nächsten BeitragZum vorigen BeitragZum erstem Beitrag2019-12-04 23:02
xiao_shi_tou_
 

Jemand schreibt:
Ich bin euch sehr dankbar, wenn ihr mir erklären könntet was ich hier flasch mache.

Gruppen
Universität/Hochschule 
Thema eröffnet von: Donauschwabe
Gruppenhomomorphismen zyklischer Gruppen  
Beitrag No.1 im Thread
Zum letzten BeitragZum nächsten BeitragZum vorigen BeitragZum erstem Beitrag2019-12-04 22:45
xiao_shi_tou_
 
\(\begingroup\)\( \DeclareMathOperator{\mer}{mer} \DeclareMathOperator{\Sht}{Sht} \DeclareMathOperator{\Ann}{Ann} \DeclareMathOperator{\Et}{\acute{E}t} \DeclareMathOperator{\et}{\acute{e}t} \DeclareMathOperator{\etaleness}{\acute{e}taleness} \newcommand{\h}{\o{h}} \newcommand{\vp}{\varphi} \newcommand{\unr}[1]{#1^{\o{un}}} \renewcommand{\H}{\o{H}} \DeclareMathOperator{\ind}{ind} \DeclareMathOperator{\etale}{\acute{e}tale} \DeclareMathOperator{\Coker}{Coker} \DeclareMathOperator{\Div}{Div} \DeclareMathOperator{\Gl}{GL} \DeclareMathOperator{\PGL}{PGL} \DeclareMathOperator{\dom}{dom} \DeclareMathOperator{\PSL}{PSL} \DeclareMathOperator{\SL}{SL} \DeclareMathOperator{\Res}{Res} \DeclareMathOperator{\equi}{equi} \DeclareMathOperator{\Hecke}{Hecke} \DeclareMathOperator{\Aut}{Aut} \DeclareMathOperator{\Jac}{Jac} \DeclareMathOperator{\GL}{GL} \DeclareMathOperator{\HF}{HF} \DeclareMathOperator{\HS}{HS} \DeclareMathOperator{\Ker}{Ker} \DeclareMathOperator{\trdeg}{trdeg} 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2019-12-04 22:04 - Donauschwabe im Themenstart schreibt:
Guten Abend,

ich möchte/soll alle Gruppenhomomorphismen \(\IZ/12\IZ\to \IZ/27\IZ\) bestimmen.

Drei fielen mir schnell ein:

\(\phi_1:\ g\mapsto g\)
\(\phi_2:\ g\mapsto g^3\)
\(\phi_3:\ g\mapsto e\)



Danke für jede Hilfe  :-)
Hi Donauschwabe.
Ich würde die Homomorphismen hier additiv schreiben, da die Gruppen additiv sind.

Die Abbildung $g\mapsto g$ ist kein Homomorphismus.
EDIT: $g\mapsto g^3$ ist auch kein Homomorphismus.

Die Homomorphismen $\Z/{12}\to \Z/{27}$ entsprechen $1-1$ den Homomorphismen $\Z\to \Z/{27}$ die $12$ auf $0$ schicken.

Das ist genau die Aussage des Homomorphiesatzes.

Diese entprechen $1-1$ den Elementen $z\in \Z/{27}$ mit $12z=0$.
Das liegt daran, dass sich die Gruppe $\Z$ durch ein Element erzeugen lässt. Folglich ist ein Homomorphismus auf $\Z$ schon durch das Bild auf dem Erzeuger eindeutig bestimmt. Für Details empfehle ich dir den Artikel den ich unter "PS" angefügt habe.

Unter dieser Bijektion entspricht ein Element $z\in \Z/{27}$ mit $12z=0$ dem Homomorphismus der durch $1\mapsto z$ eindeutig festgelegt ist.*
Ein Element $x\in \Z/{27}$ ist $=0$ genau dann, wenn $27\mid x$ gilt.



XST
PS: Schau mal  hier rein.
* Ich habe hier als Erzeuger von $\Z$ die $1$ genommen. $-1$ hätte es ber auch getan.
\(\endgroup\)

Rätsel und Knobeleien (Knobelecke)
  
Thema eröffnet von: MartinN
* 44/53  
Beitrag No.16 im Thread
Zum letzten BeitragZum nächsten BeitragZum vorigen BeitragZum erstem Beitrag2019-12-01 10:17
xiao_shi_tou_
 
\(\begingroup\)\( \DeclareMathOperator{\mer}{mer} \DeclareMathOperator{\Sht}{Sht} \DeclareMathOperator{\Ann}{Ann} \DeclareMathOperator{\Et}{\acute{E}t} \DeclareMathOperator{\et}{\acute{e}t} \DeclareMathOperator{\etaleness}{\acute{e}taleness} \newcommand{\h}{\o{h}} \newcommand{\vp}{\varphi} \newcommand{\unr}[1]{#1^{\o{un}}} \renewcommand{\H}{\o{H}} \DeclareMathOperator{\ind}{ind} \DeclareMathOperator{\etale}{\acute{e}tale} \DeclareMathOperator{\Coker}{Coker} \DeclareMathOperator{\Div}{Div} \DeclareMathOperator{\Gl}{GL} \DeclareMathOperator{\PGL}{PGL} \DeclareMathOperator{\dom}{dom} \DeclareMathOperator{\PSL}{PSL} \DeclareMathOperator{\SL}{SL} \DeclareMathOperator{\Res}{Res} \DeclareMathOperator{\equi}{equi} \DeclareMathOperator{\Hecke}{Hecke} \DeclareMathOperator{\Aut}{Aut} \DeclareMathOperator{\Jac}{Jac} \DeclareMathOperator{\GL}{GL} \DeclareMathOperator{\HF}{HF} \DeclareMathOperator{\HS}{HS} \DeclareMathOperator{\Ker}{Ker} \DeclareMathOperator{\trdeg}{trdeg} 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2019-11-30 23:39 - StrgAltEntf in Beitrag No. 12 schreibt:
Mal eine Frage. Alle bisherigen Vorschläge benötigten sechs Stammbrüche oder mehr. Ist es eigentlich klar, dass es nicht mit fünf geht? Das wäre doch auch ein schöner Rekord.

Grüße
StrgAltEntf

$$\frac{44}{53}=\frac{1}{2} + \frac{1}{5} + \frac{1}{9} + \frac{1}{53} + \frac{1}{4770}$$
\(\endgroup\)

Rätsel und Knobeleien (Knobelecke)
  
Thema eröffnet von: MartinN
* 44/53  
Beitrag No.15 im Thread
Zum letzten BeitragZum nächsten BeitragZum vorigen BeitragZum erstem Beitrag2019-12-01 10:06
xiao_shi_tou_
 
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$$\frac{44}{53}=\frac{1}{2}+\frac{1}{4}+\frac{1}{22}+\frac{1}{53}+\frac{1}{159}+\frac{1}{212}+\frac{1}{318}+\frac{1}{583}$$
Ich glaube, den Rekord von Kay_S mit $371$ wird so schnell niemand knacken :D.
\(\endgroup\)

Polynome
Universität/Hochschule 
Thema eröffnet von: IVmath
Beweis Komposition von Polynomen  
Beitrag No.21 im Thread
Zum letzten BeitragZum nächsten BeitragZum vorigen BeitragZum erstem Beitrag2019-11-30 23:57
xiao_shi_tou_
J
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Hi IVmath.
Du hast gefragt welche Polynomfunktionen $p\colon \C\to \C$ surjektiv sind.
Der Hinweis von Kezer war den Fundamentalsatz der Algebra anzuwenden.

Man kann diesen direkt auf Wikipedia nachschlagen:

"Der (Gauß-d’Alembertsche) Fundamentalsatz der Algebra besagt, dass jedes nicht konstante Polynom im Bereich der komplexen Zahlen mindestens eine Nullstelle besitzt. Dabei können die Koeffizienten des Polynoms beliebige komplexe Zahlen sein"

Nehmen wir also mal an, dass dieser Satz stimmt, ohne uns an einem Beweis aufzuhalten.

Sei nun $p\colon \C\to \C$ eine Polynomfunktion die durch ein Polynom $p(X)\in \C[X]$ gegeben ist. Was bedeutet es, dass $p$ surjektiv ist?
Das bedeutet dass es zu jedem $z\in \C$ ein $z'\in\C$ gibt mit $z=p(z')$.


Es ist offensichtlich, dass eine konstante Funktion nicht surjektiv sein kann, denn eine konstante Funktion kann nur einen Wert annehmen, aber eine surjektive Funktion muss jede Komplexe Zahl als Wert annehmen!
Nehmen wir also an, dass $p$ nicht konstant ist.
Dann ist insbesondere das Polynom auch nicht konstant.

Damit man den Fundamentalsatz der Algebra anwenden kann muss man erstmal die Aussage $z=p(z')$ etwas umformulieren, damit man eine Aussage über die Existenz von Nullstellen hat. (Verknüpfe immer das Gegebene mit dem Gesuchten.)
Die Aussage $z=p(z')$ ist äquivalent zu $p(z')-z=0$.
Das bedeutet, dass das Polynom $f(X)=p(X)-z$ genau dann eine Nullstelle besitzt, wenn es ein $z'\in\C$ gibt mit $p(z')=z$ gibt.

Wenden wir nun den Fundamentalsatz der Algebra an:
Wenn die Polynomfunktion $p$ durch ein nicht konstantes Polynom $p(X)$ gegeben ist, dann ist die Funktion $p$ surjektiv, da es nach dem Fundamentalsatz der Algebra, für jede komplexe Zahl $z$ eine Nullstelle des nicht konstanten Polynoms $p(X)-z$ in $\C$ gibt.
Nimmt man umgekehrt an, dass eine Polynomfunktion surjektiv ist, dann ist sie insbesondere nicht konstant.
Wir haben also bewiesen, dass eine Polynomfunktion $p\colon \C\to \C$ genau dann surjektiv ist, wenn sie nicht konstant ist.

Ein paar ehrlich gemeinte Worte
Wie in jeder anderen Disziplin ist es auch (vor allem!) in der Mathematik unmöglich eine schwierige Aufgabe zu lösen, wenn man nicht einmal die Grundlagen nicht kennt.

Noch kein Mensch ist vom Himmel gefallen und konnte auf Anhieb Beethovens Sonaten spielen und noch kein Mensch konnte auf Anhieb eine Fremdsprache sprechen und noch kein Mensch ist als Profi Fußballer vom Himmel gefallen.

Im Gegenteil: Von allen Experten und Menschen die ich persönlich kenne die etwas geschafft haben, was andere nicht geschafft haben weiß ich, dass das mitunter wesentlich daran lag, dass sie die Grundlagen ihrer Disziplin sehr viel besser kannten als ich zum Beispiel (das gebe ich gerne und offen zu) oder als der Großteil aller Mitstreiter auf ihrem Gebiet und, dass sie ein starkes Interesse für die Tätigkeit mitbrachten.

Bei dir kann man weder Interesse für die Mathematik als ganzes noch für die Grundlagen erkennen.

Das ist meiner Einschätzung nach wohl der Hauptgrund, dass man dir hier nicht mehr so viel antwortet. Sobald du echtes Interesse und bereit bist dich in die Materie einzuarbeiten wird man dir hier sicher gerne dabei helfen. Wenn du aber weiterhin nur eine Vermutung beweisen willst und sonst nichts mit der Mathematik zu tun haben willst, dann wird sich wahrscheinlich kaum jemand für dein Projekt begeistern lassen.

Viele Grüße
XST


 

\(\endgroup\)

Logik, Mengen & Beweistechnik
Universität/Hochschule 
Thema eröffnet von: rapiz
Beweise verbessern  
Beitrag No.2 im Thread
Zum letzten BeitragZum nächsten BeitragZum vorigen BeitragZum erstem Beitrag2019-11-29 17:05
xiao_shi_tou_
 
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2019-11-29 13:32 - rapiz im Themenstart schreibt:

$$
\begin{array}{l}{\text { Satz. Es gibt unendlich viele Primzahlen. }} \\ {\text { Beweis. Durch Wiederspruch: endlich viele Primzahlen, } \Rightarrow \text { dann ist } p \text { die gröBte Primzahl. Jede Zahl } n>p \text { ist, }} \\ {\text { durch 1 }<q \leq p \text { teilbar. Betrachte nun } n:=p !+1 \text { . Trivial } n>p \text { , und nicht } n \text { teilbar durch } q \text { (da } \forall q \text { gilt } p !=n-1 \text { ist }} \\ {\text { durch } q \text { teilbar } \rightarrow p !+1=n \text { nicht). Widerspruch und } p \text { gibt es nicht. }}\end{array}
$$
"Widerspruch" und "größte Primzahl" sollte es wohl heißen.
\(\endgroup\)

Zahlen - Darstellbarkeit
Universität/Hochschule 
Thema eröffnet von: rapiz
Beweis für jede Wurzel einer Primzahl ist irrational  
Beitrag No.20 im Thread
Zum letzten BeitragZum nächsten BeitragZum vorigen BeitragZum erstem Beitrag2019-11-29 16:50
xiao_shi_tou_
 
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$a^2$ und $b^2p$ haben nicht die gleichen Teiler. $a^2$ ist eine Quadratzahl und kann daher nur gerade viele $p$ als Faktor enthalten, während die Anzahl der $p$ in $b^2p$ ungerade sein muss.  

Man kann von vornherein annehmen, dass $a,b$ teilerfremd sind, oder dass $a$ minimal ist, aber man muss das nicht.
[Die Antwort wurde nach Beitrag No.17 begonnen.]
\(\endgroup\)

Algebraische Geometrie
Universität/Hochschule 
Thema eröffnet von: xiao_shi_tou_
Gerben über Schemata  
Beitrag No.4 im Thread
Zum letzten BeitragZum nächsten BeitragZum vorigen BeitragZum erstem Beitrag2019-11-28 23:24
xiao_shi_tou_
 
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2019-11-28 19:09 - Triceratops in Beitrag No. 3 schreibt:
2019-11-28 15:44 - xiao_shi_tou_ in Beitrag No. 2 schreibt:
Mich würde noch interessieren, ob man Gerben durch universelle Eigenschaften Charakterisieren kann!?

Kannst du diese Frage präzisieren?

Gerben sind ja Stacks mit einer gewissen Eigenschaft, siehe stacks.math.columbia.edu/tag/06NY . Diese Definition ist aber ganz offensichtlich keine universelle Eigenschaft.
Hi Triceratops.
Ich dachte an eine Eigenschaft der Art:
Ein morphismus von stacks $\c{F}\to \c{G}$ über einem Situs $\c{C}$ ist eine $\Gm$-Gerbe (oder allgemeiner für eine andere Gruppe), genau dann, wenn die Universelle Eigenschaft... erfüllt ist.

Vielleicht meine ich auch etwas wie einen Quotienten stack, aber damit bin ich noch nicht so vertraut. Ich müsste mich noch ein bisschen mehr mit den Begriffen beschäftigen, vielleicht klärt sich meine Frage dann von selbst.

Ich habe jetzt aber ein anderes interessantes Resultat in einem Paper von V.Lafforge gefunden:
"Every quotient of an algebraic variety by a finite étale group scheme is a Deligne-Mumford stack."

Ich glaube, dass aus dieser Aussage zusammen mit der Darstellbarkeit des Picard Funktors folgen könnte, dass der Picard stack ein DM-stack ist. Dem will ich jetzt mal auf den Grund gehen.

Viele Grüße
XST
 

\(\endgroup\)

Kategorientheorie
Universität/Hochschule 
Thema eröffnet von: xiao_shi_tou_
Stack nicht darstellbar durch Schema wegen nicht-trivialen Automorphismen  
Beitrag No.2 im Thread
Zum letzten BeitragZum nächsten BeitragZum vorigen BeitragZum erstem Beitrag2019-11-28 15:47
xiao_shi_tou_
J
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2019-11-27 20:49 - Triceratops in Beitrag No. 1 schreibt:
Das Argument ist OK, aber du hast ein paar Tippfehler drin (passen X,Y überall?).

Gemeint ist vielleicht ein $2$-Endomorphismus in der $2$-Kategorie der Stacks des $1$-Morphismus $X : \{*\} \to \mathcal{F}$, was dann dasselbe wie ein Endomorphismus von $X \in \mathcal{F}$ ist.
Danke.
Ich habe es korrigiert. Ich hatte zuerst zwei $X$ und habe eins dann umbenannt.
Ja, das wäre natürlich das gleiche wie ein Endomorphismus. Darauf bin ich nicht gekommen es so zu sehen. Dennoch sehe ich beweistechnisch keinen Grund, warum man die Aussage so formulieren sollte. Da das Skript jetzt aber schon mehrere Fehler aufwies denke ich, dass es auch ein Fehler sein könnte.
\(\endgroup\)

Algebraische Geometrie
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Beitrag No.2 im Thread
Zum letzten BeitragZum nächsten BeitragZum vorigen BeitragZum erstem Beitrag2019-11-28 15:44
xiao_shi_tou_
 
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2019-11-27 20:44 - Triceratops in Beitrag No. 1 schreibt:
Ich kenne mich mit dem Thema nicht wirklich gut aus, aber ich schätze, dass
1) die beiden Aussagen nicht inhaltlich voneinander abhängen,
2) die Aussage, dass $\mathcal{Pic}$ ein DM-Stack ist, viel einfacher als die Darstellbarkeit von $\mathrm{Pic}$ zu beweisen ist. Hast du dir die Beweise einmal angesehen?

Hallo Triceratops.
Ja, ich habe mir die Beweise angesehen, aber noch nicht sehr sorgfältig. Der Beweis zur Darstellbarkeit des Picard Funktors ist wirklich nicht so einfach (ich habe den Beweis im Buch über Neron Modelle von Bosch vorliegen). Wenn überhaupt würde also eher aus der Darstellbarkeit des Picard Funktors folgen, dass $\cPic_{X/S}$ ein DM-stack ist, aber ich sehe nicht, wie. Wenn es nicht möglich ist, dann würde ich eben den Beweis in stacks Projekt nehmen.

Ich dachte, dass eine Verbindung bestehen könnte, weil die Voraussetzungen an den Strukturmorphismus $X\to S$ sehr ähnlich sind.
Mich würde noch interessieren, ob man Gerben durch universelle Eigenschaften Charakterisieren kann!? Ich habe bisher nur die übliche Definition wie in stacks project gefunden.

Vielen Dank für die Hilfsbereitschaft
Viele Grüße
XST


 
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Beitrag No.4 im Thread
Zum letzten BeitragZum nächsten BeitragZum vorigen BeitragZum erstem Beitrag2019-11-28 12:55
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Häh? Das mach ich doch schon immer so :-D.

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xiao_shi_tou_
 
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Hallo zusammen.
Vor kurzem habe ich die Begriffe "Gerbe" und "Algebraischer Stack (DM-Stack)" kennengelernt und habe nun eine Frage zum Thema.
Es gibt ja den Picard stack $\cPic_{X/S}\colon (\mathbf{Sch}/S)_{fppf}^{op}\to \mathbf{Grpd}$ und den relativen Picard Funktor $\Pic_{X/S}\colon (\mathbf{Sch}/S)_{fppf}^{op}\to \mathbf{Sets}$.

Man kann dann zeigen, dass der Morphismus
$$\cPic_{X/S}\to \Pic_{X/S}$$,
welcher ein Geradenbündel $\lineb$ auf seine Isomorphie Klasse $[\lineb]$ schickt eine $\mathbb{G}_m$-Gerbe ist. Objekte in $\cPic_{X/S}$ haben $\Gm$ als Automorphismengruppe.

Man sieht auch ein, dass der Picard Funktor als $fppf$-Garbifizierung von
$T\mapsto \cPic_{X/S}\mid_T/{\cong}$ gegeben ist.

Der Picard stack selbst kann nicht als Schema darstellbar sein, da es nicht-triviale Automorphismen gibt, doch der relative Picard Funktor ist unter gewissen Voraussetzungen an $X\to S$ bekanntlich durch ein Schema darstellbar (Grothendieck).

Unter passenden Voraussetzungen an $X\to S$ kann man aber zeigen, dass $\cPic_{X/S}$ ein DM-Stack ist, also insbesondere bedeutet das ja, dass jeder Morphismus $T\to \cPic_{X/S}$ von einem Schema $T$ darstellbar ist und dass es einen $\etale n$ surjektiven Atlas $U\to \cPic_{X/S}$ gibt.

Ich verstehe das so, dass eine DM-stack das beste ist was man als Ersatz für ein Schema bekommen kann, wenn der stack nicht darstellbar ist und das der Atlas sozusagen die beste Antwort auf die Frage ist, wie man aus dem Stack ein Schema machen könnte, wobei mir nicht klar ist, ob man in der Praxis mit dem Stack selbst arbeitet (denn die Theorie der algebraischen stacks ist ja sehr ausgereift) oder ob man immer mit dem Atlas arbeitet, beziehungsweise, welche Rolle dieser Atlas spielt?

Mich interessiert nun hauptsächlich, wie die beiden Aussagen:
$\bul$ $\Pic_{X/S}$ ist durch ein Schema darstellbar.
$\bul$ $\cPic_{X/S}$ ist ein DM-stack
miteinander zusammenhängen.
Sind sie äquivalent, oder kann man wenigstens eins aus dem anderen folgern?

Wenn ich zum Beispiel die Darstellbarkeit von $\Pic_{X/S}$ annehme, dann wäre ja eine Frage, was denn dann der Atlas ist.
Klar ist ja schonmal, dass es einen Morphismus $U\to \cPic_{X/S}\to \Pic_{X/S}=h_{\underline{\Pic}_{X/S}}$ geben müsste, also einen Morphismus $U\to \underline{\Pic}_{X/S}$.
Ich habe versucht $\hom(-,\cPic)$ mit $\hom_S(-,\underline{\Pic}_{X/S})$ in Verbindung zu bringen, denn dann könnte man zum Beispiel die Identität in $\Pic_{X/S}(\underline{\Pic}_{X/S})=\hom_S(\underline{\Pic}_{X/S},\underline{\Pic}_{X/S})$ auf einen Morphismus $U\colon=\underline{\Pic}_{X/S}\to \cPic_{X/S}$ schicken, der dann als Atlas fungieren soll.

Das Problem ist, dass mir nicht klar ist, wie $\hom(-,\cPic_{X/S})$ mit $\hom(-,\Pic_{X/S})$ zusammenhängt. Insbesondere ist mir nicht klar in welchen Kategorien ich die Morphismen betrachten sollte, denn man kann ja den Funktor $\Pic_{X/S}$ auch als Stack auffassen. (Gibt es so etwas wie einen linksadjungierten Funktor zum Vergissfunktor (stacks)$\to$ ($fppf$-Garben)?).

Für die kovarianten Funktoren $\hom(\cPic_{X/S},-)$ und $\hom(\underline{\Pic}_{X/S},-)$ würde ich hingegen etwas ähnliches wie $\hom(X/\sim,-)\cong \set{\psi\in \hom(X,-)}{x\sim x'\implies \psi(x)=\psi(x')}$ erwarten, da ja $\Pic_{X/S}$ aus dem stack entsteht indem man die Isomorphismen rausteilt. Aber das würde mich bei meiner Frage glaube ich nicht weiter bringen, falls es überhaupt stimmt.

Einen Beweis, dass $\cPic_{X/S}$ ein algebraischer Stack ist gibt es ja im stacks projekt, aber mir geht es hauptsächlich darum, ob und wie das mit der Darstellbarkeit des Picard Funktors zusammenhängt.

Ich hoffe ich konnte meine Frage verständlich erklären und entschuldige mich dafür, dass sie so lang geworden ist.

Viele Grüße
XST
\(\endgroup\)

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Thema eröffnet von: xiao_shi_tou_
Stack nicht darstellbar durch Schema wegen nicht-trivialen Automorphismen  
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Zum letzten BeitragZum nächsten BeitragZum vorigen BeitragZum erstem Beitrag2019-11-27 17:42
xiao_shi_tou_
J
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Hallo zusammen.
Die Aussage ist, dass ein stack, welcher ein Objekt mit einem nicht-trivialen Automorphismus besitzt nicht durch ein Schema darstellbar sein kann, weil es dann einen nicht-trivialen $2$-Morphismus gäbe.

Ich sehe ein, dass die Aussage wahr ist, aber nicht, was das mit einem $2$-Morphismus zu tun hat (von welcher $2$-Kategorie?)

Sei $Y\in \o{\Ob}(\c{F})$ ein Objekt, sagen wir über $U\in (\mathbf{Sch}/S)_{fppf}$ und sei $\psi\colon Y\sto Y$ ein nicht-trivialer Automorhpismus.
Wäre nun $\c{F}\cong \h_X$ und sei $Y'$ das Bild von $Y$ unter diesem isomorphismus in dem Stack $\h_X$. Dann wäre das Bild des automorphismus
$Y\sto Y$ ein nicht-trivialer Automorphismus in der Faserkategorie von $\h_X$ über $U$, aber diese Faserkategorie hat nur Identitäten als Morphismen.

Welcher $2$-Morphismus ist hier gemeint und von welcher $2$-Kategorie?
Viele Grüße
XST  
\(\endgroup\)

Spiel & Spaß
  
Thema eröffnet von: mire2
MP-Stilblüten etc. sammeln  
Beitrag No.1385 im Thread
Zum letzten BeitragZum nächsten BeitragZum vorigen BeitragZum erstem Beitrag2019-11-26 00:03
xiao_shi_tou_
 

2019-11-25 23:05 - StrgAltEntf in Beitrag No. 1384 schreibt:
2019-11-25 22:51 - Slash in Beitrag No. 1383 schreibt:
Ein kleiner Fakt, der gerade ganz gut passt: Es leben mehr Bakterien in und auf uns, als unser Körper an Zellen aufweist. Ein Erwachsener besteht aus 100 Billionen bzw. 100.000.000.000.000 einzelnen Zellen.

Ist ja eklig!  eek

Bestehen Bakterien nicht auch aus Zellen? Heißt das also, dass weniger als 50% von uns zu uns gehören?
Das ist erstmal eklig...

Komplexe Zahlen
Universität/Hochschule 
Thema eröffnet von: marathon
Quotient von Potenzen komplexer Zahlen  
Beitrag No.20 im Thread
Zum letzten BeitragZum nächsten BeitragZum vorigen BeitragZum erstem Beitrag2019-11-25 23:40
xiao_shi_tou_
 
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2019-11-25 04:21 - marathon im Themenstart schreibt:

nun habe ich den Zähler mit i^4
multiplizert und den Nenner ebenso  damit keine Veränderung entsteht
Brauchst du nicht. Wegen $i^4=1$ darfst du innerhalb der Klammer $(\dotsc)^4$ nach belieben mit $i$ multiplizieren. Du darfst wegen $(-1)^4=1$ auch beliebig das Vorzeichen vertauschen.
\(\endgroup\)

Komplexe Zahlen
Universität/Hochschule 
Thema eröffnet von: marathon
Quotient von Potenzen komplexer Zahlen  
Beitrag No.19 im Thread
Zum letzten BeitragZum nächsten BeitragZum vorigen BeitragZum erstem Beitrag2019-11-25 23:32
xiao_shi_tou_
 
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@Marathon
Ich würde so vorgehen:
Sei $x$ die Zahl die du ausrechnen willst.
Beachte erstmal, dass $(\sqrt{3}+i)^4=(i\sqrt{3}-1)^4=(1-i\sqrt{3})^4$ gilt.
Wenn du nun $a=1-i\sqrt{3}$ setzt dann hast du $x=16 \frac{a^4}{\cl{a}^5}$.
Setzt man nun $a=\nrm{a}e^{i\a}$
dann folgt:
$x=16 \nrm{a}^{-1}\frac{e^{4i\a}}{e^{-5i\a}}=16 \nrm{a}^{-1}e^{(4+5)i\a}= 16 \nrm{a}^{-1}e^{9i\a}$.
$\nrm{a}$ und $\a$ ausrechnen ist nicht schwer.



\(\endgroup\)

Komplexe Zahlen
Universität/Hochschule 
Thema eröffnet von: marathon
Quotient von Potenzen komplexer Zahlen  
Beitrag No.18 im Thread
Zum letzten BeitragZum nächsten BeitragZum vorigen BeitragZum erstem Beitrag2019-11-25 23:07
xiao_shi_tou_
 
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2019-11-25 22:30 - geroyx in Beitrag No. 17 schreibt:
2019-11-25 22:28 - xiao_shi_tou_ in Beitrag No. 16 schreibt:
Schönheit liegt im Auge des Betrachters.

Ja, der Standard wäre schon schön genug.

PS: Wenn man das nicht als geschlossene Gleichungskette notieren will oder kann, dann führt man zumindest sinnvolle Platzhalter ein; und greift diese wieder auf.
Dieses Rausgreifen einzelner Terme, und an denen werden dann noch irgendwelche Umformungen durchgeführt, ist grundsätzlich schlechter Stil. Das ist m.E. auch keine Geschmacksfrage.  

Was Dietmar gemacht hat ist logisch und effizient.
Er hat erstmal die Werte von $2z$ und von $\cl{z}i$ ausgerechnet und dann die Potenzen $(2z)^4$ und $(\cl{z}i)^5$.
Das ergibt insgesamt $4$ Rechnungen die er übersichtlich hingeschrieben hat. Das ist sein Stil und ich sehe hier kein Problem.

Alles in eine Gleichungskette zu schreiben kann mehr Schreibarbeit erfordern und ist bei etwas komplizierteren Beweisen gar nicht möglich.
Der Beweis der Endlichkeit der Primzahlen geht auch nicht so "Anzahl der Primzahlen$=\dotsc<\infty$".

Ganz davon abgesehen ist es völlig egal, ob dir sein Stil gefällt oder nicht, denn es ist sein Stil und es ist seine Entscheidung Dinge so aufzuschreiben wie er es möchte.

Vielen Dank
\(\endgroup\)

Komplexe Zahlen
Universität/Hochschule 
Thema eröffnet von: marathon
Quotient von Potenzen komplexer Zahlen  
Beitrag No.16 im Thread
Zum letzten BeitragZum nächsten BeitragZum vorigen BeitragZum erstem Beitrag2019-11-25 22:28
xiao_shi_tou_
 

2019-11-25 22:25 - geroyx in Beitrag No. 15 schreibt:
Ok, ich habe es erst jetzt gesehen; und entschuldige mich.

Ich plädiere trotzdem für eine schöne Notation.

[Die Antwort wurde nach Beitrag No.13 begonnen.]

Schönheit liegt im Auge des Betrachters.

Komplexe Zahlen
Universität/Hochschule 
Thema eröffnet von: marathon
Quotient von Potenzen komplexer Zahlen  
Beitrag No.13 im Thread
Zum letzten BeitragZum nächsten BeitragZum vorigen BeitragZum erstem Beitrag2019-11-25 22:23
xiao_shi_tou_
 
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2019-11-25 22:11 - geroyx in Beitrag No. 12 schreibt:
2019-11-25 22:03 - dietmar0609 in Beitrag No. 11 schreibt:
Auch wenn deine "Korrekturen" in rot erscheinen, sind sie falsch und zeigen, dass du meinen Beitrag nicht verstanden oder gelesen hast.  
..
Die Gleichheit $4\cdot e^{i\cdot \frac{\pi}{6}}=(4\cdot e^{i\cdot \frac{\pi}{6}})^4$ die du behauptest ist falsch.


\(\endgroup\)

Algebraische Geometrie
Universität/Hochschule 
Thema eröffnet von: xiao_shi_tou_
Höhere Bilder von Garben auf algebraischen Stapeln  
Beitrag No.3 im Thread
Zum letzten BeitragZum nächsten BeitragZum vorigen BeitragZum erstem Beitrag2019-11-24 19:43
xiao_shi_tou_
J
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Hallo Triceratops.
Danke für die Hilfe.
Den zweiten Link den du angegeben hast hatte ich nicht gefunden.
Wenn ich mir das zum $fppf$-Situs noch durchlese sollte alles klar sein.
Vielen Dank
PS:
$\bul$ Nächstes Mal schreibe ich den Link dazu.
$\bul$ "Stapel"$\colon\defeq$ "stack"
\(\endgroup\)
 

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